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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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The area of the region bounded by the y-axis ,$y=\cos x$ and $y=\sin x,0\leq x\leq \large\frac{\pi}{2}$ is\begin{array}{1 1}(A)\;1\sqrt 2sq.units & (B)\;(\sqrt 2+1)sq.units\\(C)\;(\sqrt 2-1)sq.units & (D)\;(2\sqrt 2-1)sq.units \end{array}

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Toolbox:
  • The area enclosed by the curve $y=f(x)$,the $x$-axis and the ordinates $x=a$ and $x=b$ is given by $\int_a^b ydx.$
  • $\sin(A-B)=2\cos\large\frac{(A+B)}{2}$$\sin\large\frac{(A-B)}{2}$
  • Principle solution of $\sin\theta=n\pi\pm(-1)^n\alpha$
  • $\int \cos \theta=\sin\theta+c$
Step 1:
We have to find the area of the region bounded by the $y$-axis,$y=\cos x$ and $y=\sin x$.
Where $x$ ranges between $0\leq x\leq \large\frac{\pi}{2}$
First to find the area,let us find the point of intersection of $f(x)=\sin x$ and $g(x)=\cos x$
At the point of intersection $\sin x=\cos x$
$\Rightarrow \sin x=\sin(\large\frac{\pi}{2}$$-x$)=0
Now applying $\sin(A-B)=2\cos\large\frac{(A+B)}{2}$$\sin\large\frac{(A-B)}{2}$,we get
$\;\;=2\cos \big(\large\frac{x+\Large\frac{\pi}{2}-x}{2}\big)\normalsize\sin\big(\large\frac{x-(\Large\frac{\pi}{2}-x)}{2}\big)$$=0$
$\;\;=2\cos\large\frac{\pi}{4}$$\sin(x-\large\frac{\pi}{4})$$=0$
But $\cos\large\frac{\pi}{4}=\large\frac{1}{\sqrt 2}$
$\Rightarrow 2.\large\frac{1}{\sqrt 2}$$\sin(x-\large\frac{\pi}{4})$$=0$
$\sin(x-\large\frac{\pi}{4})$$=0\Rightarrow x-\large\frac{\pi}{4}$$=0$
Step 2:
Applying the principle solution we get,
$x-\large\frac{\pi}{4}=$$n\pi$
$\Rightarrow x=\large\frac{\pi}{4},\frac{5\pi}{4},\frac{9\pi}{4}$........
But it is given $x$ values ranges between $0$ and $\large\frac{\pi}{2}$
Hence we take only $\large\frac{\pi}{4}$ and discard the 0 points.
Required area is the shaded portion shown in the fig.
The area of the shaded portion is given by $A=\int_0^{\Large\frac{\pi}{4}}(\sin x-\cos x)dx.$
Step 3:
On integrating we get,
$A=\begin{bmatrix}-\cos x-\sin x\end{bmatrix}_0^{\Large\frac{\pi}{4}}$
On applying the limits we get,
$A=[-(\cos\large\frac{\pi}{4}-$$\cos 0)-(\sin\large\frac{\pi}{4}$$-\sin 0)]$
But $\sin\large\frac{\pi}{4}=$$\cos\large\frac{\pi}{4}=\frac{1}{\sqrt 2},$$\cos 0=1$ and $\sin 0=0$
Substituting the values we get,
$\;\;\;=[-[\large\frac{1}{\sqrt 2}$$-1]-[\large\frac{1}{\sqrt 2}]]$
$\;\;\;=1-\large\frac{2}{\sqrt 2}=$$(1-\sqrt 2)$ or $\sqrt 2-1$
$A=(\sqrt 2-1)$sq.units.
Hence C is the correct option.
answered May 6, 2013 by sreemathi.v
edited May 6, 2013 by sreemathi.v
 

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